Yet of the three circle lessons that are tested on the EOY, Section 15-3 is the only one that actually appears in Chapter 15. The other two lessons are Section 13-5 on tangents to circles and yesterday's Section 11-3 on equations of circles. Therefore today's worksheet contains the only lesson that I'll include from Chapter 15.

Section 15-3 of the U of Chicago text is on the Inscribed Angle Theorem. I admit that I often have trouble remembering all of the circle theorems myself, but this one is the most important:

Inscribed Angle Theorem:

In a circle, the measure of an inscribed angle is one-half the measure of its intercepted arc.

The text divides the proof into three cases -- depending on whether the center of the circle is inside, outside, or on the inscribed angle. The easiest case occurs when the center is on the angle, and this case is used to prove the other two cases. The exact same thing occurred when we proved the Triangle Area Formula back in Chapter 8. Then, the three cases were whether the altitude was inside, outside, or aside of the triangle -- and the case when the altitude was aside of the triangle (i.e., when the triangle was a right triangle) was used to prove the other two.

I will reproduce the paragraph proof of the Inscribed Angle Theorem from the U of Chicago:

Given: Angle

*ABC*inscribed in Circle

*O*

Prove: Angle

*ABC*= 1/2 * Arc

*AC*

*Proof:*

Case I: The auxiliary segment

*AOB*is isosceles [both

*OA*and

*OB*are radii of the circle -- dw], Angle

*B*= Angle

*A*. Call this measure

*x*. By the Exterior Angle Theorem, Angle

*AOC*= 2

*x*. Because the measure of an arc equals the measure of its central angle, Arc

*AC*= 2

*x*= 2 * Angle

*B*. Solving for Angle

*B*, Angle

*B*= 1/2 * Arc

*AC*. QED Case I.

Notice that the trick here was that between the central angle (whose measure equals that of the arc) and the inscribed angle is an isosceles triangle. We saw the same thing happen in yesterday's proof of the Angle Bisector Theorem -- the angle bisector of a triangle is a side-splitter of a larger triangle, and cutting out the smaller triangle from the larger leaves an isosceles triangle behind.

Let's move onto Case II. Well, the U of Chicago almost gives us a two-column proof here, so why don't we complete it into a full two-column proof. For Case II,

*O*is in the interior of Angle

*ABC*.

Statements Reasons

1.

*O*interior

*ABC*1. Given

2. Draw ray

*BO*inside

*ABC*2. Definition of interior of angle

3. Angle

*ABC*= Angle

*ABD*+ Angle

*DBC*3. Angle Addition Postulate

4. Angle

*ABC*= 1/2 * Arc

*AD*+ 1/2 * Arc

*DC*4. Case I and Substitution

5. Angle

*ABC*= 1/2(Arc

*AD*+ Arc

*DC*) 5. Distributive Property

6. Angle

*ABC*= 1/2 * Arc

*AC*6. Arc Addition Property and Substitution

The proof of Case III isn't fully given, but it's hinted that we use subtraction rather than addition as we did in Case II. Once again, I bring up the Triangle Area Proof -- the case of the obtuse triangle involved subtracting the areas of two right triangles, whereas in the case where that same angle were acute, we'd be adding the areas of two right triangles.

The text mentions a simple corollary of the Inscribed Angle Theorem:

Theorem:

An angle inscribed in a semicircle is a right angle.

The text motivates the study of inscribed angles by considering camera angles and lenses. According to the text, a normal camera lens has a picture angle of 46 degrees, a wide-camera lens has an angle of 118 degrees, and a telephoto lens has an angle of 18. I briefly mention this on my worksheet. But a full consideration of camera angles doesn't occur until the next section of the text, Section 15-4 -- but we're only really doing Section 15-3.

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